Solve 1/x^2-5/x-14=0

$$\frac { 1 } { x ^ { 2 } } - \frac { 5 } { x } - 14 = 0$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{1}{7}\approx 0.142857143$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x^{2},x.

$$1-x\times 5+x^{2}\left(-14\right)=0$$

Multiply -1 and 5 to get -5.

$$1-5x+x^{2}\left(-14\right)=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-14x^{2}-5x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -14x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-14=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=2 b=-7$$

Rewrite -14x^{2}-5x+1 as \left(-14x^{2}+2x\right)+\left(-7x+1\right).

$$\left(-14x^{2}+2x\right)+\left(-7x+1\right)$$

Factor out 2x in -14x^{2}+2x.

$$2x\left(-7x+1\right)-7x+1$$

Factor out common term -7x+1 by using distributive property.

$$\left(-7x+1\right)\left(2x+1\right)$$

To find equation solutions, solve -7x+1=0 and 2x+1=0.

$$x=\frac{1}{7} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x^{2},x.

$$1-x\times 5+x^{2}\left(-14\right)=0$$

Multiply -1 and 5 to get -5.

$$1-5x+x^{2}\left(-14\right)=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-14x^{2}-5x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -14 for a, -5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-14\right)}}{2\left(-14\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-14\right)}}{2\left(-14\right)}$$

Multiply -4 times -14.

$$x=\frac{-\left(-5\right)±\sqrt{25+56}}{2\left(-14\right)}$$

Add 25 to 56.

$$x=\frac{-\left(-5\right)±\sqrt{81}}{2\left(-14\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-5\right)±9}{2\left(-14\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±9}{2\left(-14\right)}$$

Multiply 2 times -14.

$$x=\frac{5±9}{-28}$$

Now solve the equation x=\frac{5±9}{-28} when ± is plus. Add 5 to 9.

$$x=\frac{14}{-28}$$

Reduce the fraction \frac{14}{-28} to lowest terms by extracting and canceling out 14.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{5±9}{-28} when ± is minus. Subtract 9 from 5.

$$x=-\frac{4}{-28}$$

Reduce the fraction \frac{-4}{-28} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{7}$$

The equation is now solved.

$$x=-\frac{1}{2} x=\frac{1}{7}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}, the least common multiple of x^{2},x.

$$1-x\times 5+x^{2}\left(-14\right)=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-x\times 5+x^{2}\left(-14\right)=-1$$

Multiply -1 and 5 to get -5.

$$-5x+x^{2}\left(-14\right)=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-14x^{2}-5x=-1$$

Divide both sides by -14.

$$\frac{-14x^{2}-5x}{-14}=-\frac{1}{-14}$$

Dividing by -14 undoes the multiplication by -14.

$$x^{2}+\left(-\frac{5}{-14}\right)x=-\frac{1}{-14}$$

Divide -5 by -14.

$$x^{2}+\frac{5}{14}x=-\frac{1}{-14}$$

Divide -1 by -14.

$$x^{2}+\frac{5}{14}x=\frac{1}{14}$$

Divide \frac{5}{14}, the coefficient of the x term, by 2 to get \frac{5}{28}. Then add the square of \frac{5}{28} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{14}x+\left(\frac{5}{28}\right)^{2}=\frac{1}{14}+\left(\frac{5}{28}\right)^{2}$$

Square \frac{5}{28} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{14}x+\frac{25}{784}=\frac{1}{14}+\frac{25}{784}$$

Add \frac{1}{14} to \frac{25}{784} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{14}x+\frac{25}{784}=\frac{81}{784}$$

Factor x^{2}+\frac{5}{14}x+\frac{25}{784}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{28}\right)^{2}=\frac{81}{784}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{28}\right)^{2}}=\sqrt{\frac{81}{784}}$$

Simplify.

$$x+\frac{5}{28}=\frac{9}{28} x+\frac{5}{28}=-\frac{9}{28}$$

Subtract \frac{5}{28} from both sides of the equation.

$$x=\frac{1}{7} x=-\frac{1}{2}$$